# How do you write the equations for half-cell reactions?

Mar 12, 2017

Here's how I do it.

#### Explanation:

Example 1

Suppose you have the half-reaction

$\text{Cu(s)" rarr "Cu"^"2+""(aq)}$

You balance charge by adding electrons to the side that has the excess positive charge.

Here, you have two positive charges on the right and none on the left.

To balance charge, you must add two electrons to the right hand side:

$\text{Cu(s)" rarr "Cu"^"2+""(ag)" + 2"e"^"-}$

Now you have a net charge of zero on each side of the equation, so the charge is balanced.

Example 2

Suppose you have the half reaction

$\text{Ag"^"+""(s)" rarr "Ag""(s)}$

Here, you have one positive charge on the left and none on the right.

To balance charge, you must add one electron to the left hand side:

$\text{Ag"^"+""(s)" + "e"^"-" rarr "Ag""(s)}$

Now you have a net charge of zero on each side of the equation, and the charge is balanced.

Example 3

Suppose you have the half reaction

$\text{Cl"_2"(g)" rarr "2Cl"^"-""(aq)}$

Here, you have two negative charges on the right and none on the left.

To balance charge, you still add electrons to the more positive side (the left).

You add two electrons to the left hand side:

$\text{Cl"_2"(g)" + "2e"^"-" rarr "2Cl"^"-""(aq)}$

Now you have a net charge of 2- on each side of the equation, and the charges are balanced.

Example 4

Suppose you have the half reaction

${\text{MnO"_4^(-) rarr "Mn}}^{2 +}$

Now it's not so simple because we don't have the same number of elements on each side of the reaction. In cases such as these, there are four general steps we follow:

$1.$ Balance all non-oxygen and hydrogen elements.
$2.$ Balance the oxygen atoms using $\text{H"_2"O}$.
$3.$ Balance the hydrogen atoms using ${\text{H}}^{+}$.
$4.$ Balance the charges using ${\text{e}}^{-}$.

In the reaction, $\text{Mn}$ is balanced, so we must balance the oxygens. We have four on one side, so we need 4 $\text{H"_2"O}$ molecules to balance the equation

$\text{MnO"_4^(-)rarr "Mn"^(2+) + 4"H"_2"O}$

We now have the oxygen atoms balanced, so we need to balance the hydrogens. We have $8$ on one side, so we need $8$ ${\text{H}}^{+}$

$\text{MnO"_4^(-)+8"H"^+rarr "Mn"^(2+) + 4"H"_2"O}$

Now to balance the charges. On the left, the overall charge is $7$ and on the right, it's $2$. So we need $5$ electrons to balance the charges.

$\text{MnO"_4^(-)+8"H"^+ +5e^(-)rarr "Mn"^(2+) + 4"H"_2"O}$